LeetCode Implement Queue using Stacks

LeetCode Implement Queue using Stacks

Implement the following operations of a queue using stacks.

  • push(x) -- Push element x to the back of queue.
  • pop() -- Removes the element from in front of queue.
  • peek() -- Get the front element.
  • empty() -- Return whether the queue is empty.

Notes:

  • You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
  • Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
  • You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

本题要求用堆栈来实现队列的功能。因为堆栈是后进先出,而队列是先进先出,顺序正好相反。比如进入队列的顺序是1,2,3,但是进到堆栈stk1之后,从栈顶到栈底的顺序就变成了3,2,1,此时如果要实现队列出队的效果,就必须借助另一个堆栈stk2,把stk1的内容压入到stk2,这样从栈顶到栈底的顺序就和队列一样了,是1,2,3。

所以本题借助两个堆栈stk1和stk2,stk1就是常规的数据来了就压栈,当需要实现队列的pop效果时,就借助stk2倒腾一下,如果又要实现队列的push效果时,又要把stk2的数据捣腾回stk1。

总的来说,我们使用懒人规则,每次直到调用某个操作,且当前stk*不能满足要求时,才捣腾stk1和stk2。

完整代码如下:

class MyQueue {
private:
	stack<int> stk1, stk2; // 输入来源1,2,3,stk1从栈底到栈顶的顺序是1,2,3,stk2从栈底到栈顶的顺序是3,2,1
public:
	/** Initialize your data structure here. */
	MyQueue() {

	}

	/** Push element x to the back of queue. */
	void push(int x) {
		while (!stk2.empty()) {
			stk1.push(stk2.top());
			stk2.pop();
		}
		stk1.push(x);
	}

	/** Removes the element from in front of queue and returns that element. */
	int pop() {
		while (!stk1.empty()) {
			stk2.push(stk1.top());
			stk1.pop();
		}
		int top = stk2.top();
		stk2.pop();
		return top;
	}

	/** Get the front element. */
	int peek() {
		while (!stk1.empty()) {
			stk2.push(stk1.top());
			stk1.pop();
		}
		return stk2.top();
	}

	/** Returns whether the queue is empty. */
	bool empty() {
		return stk1.empty() && stk2.empty();
	}
};

本代码提交AC,用时3MS。

二刷。其实push的时候可以不需要把stk2的数据倒腾回stk1,stk2的数据始终是队列头的,stk1的数据始终是队列尾的,代码如下:

class MyQueue {
private:
    stack<int> stk1, stk2;
    void adjust() {
        while(!stk1.empty()) {
            stk2.push(stk1.top());
            stk1.pop();
        }
    }
public:
    /** Initialize your data structure here. */
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        stk1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        if(stk2.empty()) adjust();
        int ans = stk2.top();
        stk2.pop();
        return ans;
    }
    
    /** Get the front element. */
    int peek() {
        if(stk2.empty()) adjust();
        int ans = stk2.top();
        return ans;
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return stk1.empty() && stk2.empty();
    }
};

本代码提交AC,用时3MS。

One thought on “LeetCode Implement Queue using Stacks

  1. Pingback: LeetCode Implement Stack using Queues | bitJoy > code

Leave a Reply

Your email address will not be published. Required fields are marked *